两个有序链表的合并【不改变原链表】

因为要求不改变原链表,所以需要用申请内存来产生新的链表。
#include <stdlib.h>
#include <stdio.h>
#include <string.h>

typedef struct node {
    int v;
    struct node *next;
}Node;

Node *merge(Node *list1, Node *list2)
{
    Node *list = NULL, *node = NULL, *tmp = NULL;

    if(NULL == list1)
    {
        return list2;
    }
    if(NULL == list2)
    {
        return list1;
    }

    list = (Node *)malloc(sizeof(Node));
    list->next = NULL;
    if(list1->v < list2->v)
    {
        list->v = list1->v;
        list1 = list1->next;
    }
    else
    {
        list->v = list2->v;
        list2 = list2->next;
    }
    node = list;
    while(list1 && list2)
    {
        tmp = (Node *)malloc(sizeof(Node));
        memset(tmp, 0, sizeof(Node));

        if(list1->v < list2->v)
        {
           tmp->v = list1->v;

           list1 = list1->next;
        }
        else if(list1->v == list2->v)
        {
           tmp->v = list1->v;

           list1 = list1->next;
           list2 = list2->next;
        }
        else
        {
           tmp->v = list2->v;

           list2 = list2->next;
        }
        node->next = tmp;
        node = node->next;
    }

    if(NULL == list1 && NULL == list2)
    {
        return list;
    }
    else if(NULL == list1)
    {
        while(list2)
        {
            tmp = (Node *)malloc(sizeof(Node));
            memset(tmp, 0, sizeof(Node));

            tmp->v = list2->v;
            list2 = list2->next;

            node->next = tmp;
            node = node->next;
        }
    }
    else
    {
        while(list1)
        {
            tmp = (Node *)malloc(sizeof(Node));
            memset(tmp, 0, sizeof(Node));

            tmp->v = list1->v;
            list1 = list1->next;

            node->next = tmp;
            node = node->next;
        }
    }

    return list;
}

void free_list(Node *list)
{
    Node *tmp = NULL;
    while(list)
    {
        tmp = list->next;
        free(list);
        list = tmp;
    }
}

void print_list(Node *list)
{
    while(list)
    {
        printf("%u ", list->v);
        list = list->next;
    }
    printf("\n");
}

int main(int argc, char *argv[])
{
    Node *list1 = NULL;
    Node *list2 = NULL;
    Node *list = NULL;
    int len, i;


    printf("Please input the len of list1:");
    scanf("%u", &len);
    list1 = (Node *)malloc(len * sizeof(Node));
    for(i=0; i<len; i++)
    {
        scanf("%u", &list1[i].v);
        if(i == len - 1)
        {
            list1[i].next = NULL;
        }
        else
        {
            list1[i].next = &list1[i+1];
        }
    }

    printf("Please input the len of list2:");
    scanf("%u", &len);
    list2 = (Node *)malloc(len * sizeof(Node));
    for(i=0; i<len; i++)
    {
        scanf("%u", &list2[i].v);
        if(i == len - 1)
        {
            list2[i].next = NULL;
        }
        else
        {
            list2[i].next = &list2[i+1];
        }
    }

    print_list(list1);
    print_list(list2);

    list = merge(list1, list2);
    print_list(list);

    free(list1);
    free(list2);
    free_list(list);

    return 0;
}

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转载自blog.csdn.net/xqjcool/article/details/52069270